Find a matrix $B$ such that $B^* B=A$ for a given Hermitian $A$ Let
$$A=\left[
                           \begin{array}{ccc}
                             4 & 0 & 0 \\
                             0 & 1 & i \\
                             0 & -i & 1 
                           \end{array}
                         \right].$$
Find a matrix $ B $ such that $B^*B$ =$A$
(star means conjugate transpose of $B$).
I think that $A$ is hermitian, and so $A^* =A$
Also, We can edit this as $BB^*=A$.
But I couldn't solve it completely.
 A: First of all, you can't take the Hermitian of a product of matrices like that.
$$(B^*B)^*=B^*B\neq BB^* $$
You can examine the $4$ and $\left(
\begin{array}{cc}
 1 & i \\
 -i & 1 \\
\end{array}
\right)$ separately since the off diagonal elements are zero. I was able to simply guess which matrix would give  $\left(
\begin{array}{cc}
 1 & i \\
 -i & 1 \\
\end{array}
\right)$ when multiplied with its conjugate transpose.
$$B=\left(
\begin{array}{ccc}
 2 & 0 & 0 \\
 0 & 0 & 0 \\
 0 & -i & 1 \\
\end{array}
\right)$$
A: Let $A$ be an $n\times n$ Hermitian matrix. There must therefore be an orthonormal basis of eigenvectors corresponding to real eigenvalues. Write this as
$$A = \sum_k \lambda_k P_{v_k},$$
where $\lambda_k\in\mathbb R$ and $P_{v_k}\equiv v_k v_k^*$ is the orthoprojection onto the vector $v_k$.
Let $B$ be some $m\times n$ matrix whose SVD reads
$$B = \sum_k s_k (u_k w_k^*),$$
for some $s_k\ge0$ and $u_k\in \mathbb C^m, w_k\in\mathbb C^m$ and $\{u_k\}_k,\{w_k\}_k$ sets of orthonormal vectors.
Then, $B^* B$ reads
$$B^* B = \sum_k s_k^2 (w_k w_k^*) = \sum_k s_k^2 P_{w_k}.$$
If we want $B^* B=A$, we thus need $\sum_k\lambda_k P_{v_k}=\sum_k s_k^2 P_{w_k}$.
Being $\{v_k\}_k,\{w_k\}_k$ both sets of orthonormal vectors, this is only possible if, up to some relabeling, $\lambda_k=s_k^2$.
If $A$ (and thus $B^* B$) is non-degenerate, then we can also immediately conclude that $P_{v_k}=P_{w_k}$. More generally, there might be degenerate eigenvalues. In this case, the uniqueness of the spectral decomposition of a matrix tells us that the eigenspaces of $A$ and $B^* B$ corresponding to the same eigenvalue must be the same.
In conclusion, if $B^*B=A$ for $A$ Hermitian, then $B$ must have the form
$$B = \sum_k \sqrt{\lambda_k} (w_k v_k^*),$$
for some choice of orthonormal vectors $\{w_k\}_k\subset\mathbb C^m$.
